For the reaction:
Question:

For the reaction:

2A + B → A2B

the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.

Solution:

The initial rate of the reaction is

Rate = [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2

= 8.0 × 10−9 mol−2 L2 s−1

When[A] is reduced from 0.1 mol L−1 to 0.06 mol−1, the concentration of A reacted = (0.1 − 0.06) mol L−1 = 0.04 mol L−1

Therefore, concentration of $B$ reacted $=\frac{1}{2} \times 0.04 \mathrm{~mol} L^{-1}=0.02 \mathrm{~mol} L^{-1}$

Then, concentration of B available, [B] = (0.2 − 0.02) mol L−1

= 0.18 mol L−1

After [A] is reduced to 0.06 mol L−1, the rate of the reaction is given by,

Rate = [A][B]2

= (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2

 

= 3.89 mol L−1 s−1

 

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